Hodge Dual - Explanation

Explanation

Let W be a vector space, with an inner product . For every linear function there exists a unique vector v in W such that for all w in W. The map given by is an isomorphism. This holds for all vector spaces, and can be used to explain the Hodge dual.

Let V be an n-dimensional vector space with basis . For 0 ≤ k≤ n, consider the exterior power spaces and . For each and, we have . There is, up to a scalar, only one n-vector, namely . In other words, must be a scalar multiple of for all and .

Consider a fixed . There exists a unique linear function such that for all . This is the scalar multiple mentioned in the previous paragraph. If denotes the inner product on (n–k)-vectors, then there exists a unique (n–k)-vector, say, such that for all . This (n–k)-vector is the Hodge dual of λ, and is the image of the under the canonical isomorphism between and . Thus, .

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